Question regarding Basel Problem I have the following problem as quoted in a book: $$ S=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots $$
Prove that: $$ S < 3 $$
I know that $ S $ evaluates to $ \frac{\pi^2}{6} $, so it is absolutely true. But I have no idea how to prove it. Also, the author of the book wants the proof without having the knowledge of the sum of the given infinite series. Any help would be appreciated.
 A: Let’s start with $n(n+1)\gt n^2\gt n(n-1)$. This means
$${1\over n(n+1)}\lt {1\over n^2}\lt {1\over n(n-1)}$$
Now use $1/n(n+1)=1/n-1/(n+1)$ and $1/n(n-1)=1/(n-1)-1/n$ to get assuming the series converge and because the sums telescope.
$$\begin{align}
{1\over 1}-{1\over 2}&\lt {1\over 1^2}\leq 1\\
{1\over 2}-{1\over 3}&\lt {1\over 2^2}\lt {1\over 1}-{1\over 2}\\
{1\over 3}-{1\over 4}&\lt {1\over 3^2}\lt {1\over 2}-{1\over 3}\\
&\vdots\\
{1\over n}-{1\over n+1}&\lt {1\over n^2}\lt {1\over n-1}-{1\over n}\\
&\vdots
\end{align}$$
One gets eventually
$$1\lt S\lt 2$$
A: This proof is overkill, but nevertheless fun.
For every $x\in(-1,0)$, it can be shown that $\ln(1+x)<x$. Since $-1<-1/n^2<0$ is true for every $n>1$, it follows that
$$\ln\left(1-\frac{1}{n^2}\right)<-\frac{1}{n^2}$$
This is equivalent to
$$\frac{1}{n^2}<-\ln\left(1-\frac{1}{n^2}\right)$$
from which we deduce that
\begin{align}
\sum_{n=1}^{k}\frac{1}{n^2} &= 1+\sum_{n=2}^{k}\frac{1}{n^2}\\
&< 1+\sum_{n=2}^{k}-\ln\left(1-\frac{1}{n^2}\right)\\
&= 1-\sum_{n=2}^{k}\ln\left(\frac{n^2-1}{n^2}\right)\\
&= 1-\sum_{n=2}^{k}\left[\ln\left((n+1)(n-1)\right)-\ln\left(n^2\right)\right]\\
&= 1-\sum_{n=2}^{k}\left[\ln(n+1)+\ln(n-1)-2\ln(n)\right]\\
&= 1-\sum_{n=2}^{k}\left[\ln(n+1)-\ln(n)+\ln(n-1)-\ln(n)\right]\\
&= 1-\sum_{n=2}^{k}\left[\ln(n+1)-\ln(n)\right]-\sum_{n=2}^{k}\left[\ln(n-1)-\ln(n)\right]\\
&= 1-\left[\ln(k+1)-\ln(2)\right]+\sum_{n=2}^{k}\left[\ln(n)-\ln(n-1)\right]\\
&= 1-\ln(k+1)+\ln(2)+\ln(k)-\ln(2-1)\\
&= 1+\ln(2)-\left[\ln(k+1)-\ln(k)\right]\\
&= 1+\ln(2)-\ln\left(\frac{k+1}{k}\right)
\end{align}
Now, $\ln\left(\frac{k+1}{k}\right)\to 0$ as $k\to\infty$, so the sum of the series is at most $1+\ln(2)$. We can then conclude that the sum is strictly less than $2$ (and hence $3$) by leveraging the inequality $\ln(1+x)<x$ once more.
$$\sum_{n=1}^{\infty}\frac{1}{n^2}\leq 1+\ln(2) = 1+\ln(1+1) < 1+1=2$$
A: Let $$S=1+\frac14+\frac19+\frac{1}{16}+\cdots$$
Then we have: $$4S=4+1+\frac49+\frac14+\frac{4}{25}+\frac{1}{9}+\cdots\\=S+4(1+\frac19+\frac{1}{25}+\cdots)\\<S+4(1+\frac14+\frac{1}{16}+\frac{1}{36}+...)\\=S+4(1+\frac S4)$$
Then $$4S<S+4(1+\frac S4)\implies S<2$$
The third line inequality holds because $\frac14>\frac19, \frac{1}{16}>\frac{1}{25}$, etc...
